JEE Main 2026 — 24 January, Evening Shift — Question 8

Given $Q = [q_{ij}]$ with $q_{ij} = 2^{i+j-1} p_{ij}$.

$$\det(Q) = \begin{vmatrix} 2^{1+1-1}p_{11} & 2^{1+2-1}p_{12} & 2^{1+3-1}p_{13} \\ 2^{2+1-1}p_{21} & 2^{2+2-1}p_{22} & 2^{2+3-1}p_{23} \\ 2^{3+1-1}p_{31} & 2^{3+2-1}p_{32} & 2^{3+3-1}p_{33} \end{vmatrix} = \begin{vmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{vmatrix}$$

Factor out powers of 2 from each row: row 1: $2^1$, row 2: $2^2$, row 3: $2^3$.

$$\det(Q) = 2^{1+2+3} \begin{vmatrix} p_{11} & 2 p_{12} & 2^2 p_{13} \\ p_{21} & 2 p_{22} & 2^2 p_{23} \\ p_{31} & 2 p_{32} & 2^2 p_{33} \end{vmatrix} = 2^6 \cdot 2^{1+2} \begin{vmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{vmatrix}$$

Simplify: $\det(Q) = 2^{6+3} \det(P) = 2^9 \det(P)$. Given $\det(Q) = 2^{10}$, so $2^9 \det(P) = 2^{10} \Rightarrow \det(P) = 2$. For a $3 \times 3$ matrix, $\det(\operatorname{adj}(\operatorname{adj}(P))) = (\det(P))^{(3-1)^2} = (\det(P))^4$. Thus $\det(\operatorname{adj}(\operatorname{adj}(P))) = 2^4 = 16$. Hence the answer is $\boxed{B}$.